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Homeproof that $\sqrt{2}$ is irrational

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# proof that $\sqrt{2}$ is irrational

Assume that the square root of $2$ is rational. Then we can write

$\sqrt{2}=\frac{a}{b},$ |

where $a,b\in{\mathbb{N}}$ and $a$ and $b$ are relatively prime. Then $\displaystyle 2=(\sqrt{2})^{2}=\left(\frac{a}{b}\right)^{2}=\frac{a^{2}}{b^{2}}$. Thus, $a^{2}=2b^{2}$. Therefore, $2\mid a^{2}$. Since $2$ is prime, it must divide $a$. Then $a=2c$ for some $c\in{\mathbb{N}}$. Thus, $2b^{2}=a^{2}=(2c)^{2}=4c^{2}$, yielding that $b^{2}=2c^{2}$. Therefore, $2\mid b^{2}$. Since $2$ is prime, it must divide $b$.

Since $2\mid a$ and $2\mid b$, we have that $a$ and $b$ are not relatively prime, which contradicts the hypothesis. Hence, the initial assumption is false. It follows $\sqrt{2}$ is irrational.

With a little bit of work, this argument can be generalized to any positive integer that is not a square. Let $n$ be such an integer. Then there must exist a prime $p$ and $k,m\in{\mathbb{N}}$ such that $n=p^{k}m$, where $p\nmid m$ and $k$ is odd. Assume that $\sqrt{n}=a/b$, where $a,b\in{\mathbb{N}}$ and are relatively prime. Then $\displaystyle p^{k}m=n=(\sqrt{n})^{2}=\left(\frac{a}{b}\right)^{2}=\frac{a^{2}% }{b^{2}}$. Thus, $a^{2}=p^{k}mb^{2}$. From the fundamental theorem of arithmetic, it is clear that the maximum powers of $p$ that divides $a^{2}$ and $b^{2}$ are even. Since $k$ is odd and $p$ does not divide $m$, the maximum power of $p$ that divides $p^{k}mb^{2}$ is also odd. Thus, the same should be true for $a^{2}$. Hence, we have reached a contradiction and $\sqrt{n}$ must be irrational.

The same argument can be generalized even more, for example to the case of nonsquare irreducible fractions and to higher order roots.

## Mathematics Subject Classification

11J72*no label found*

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